{"id":26149,"date":"2023-06-01T10:23:06","date_gmt":"2023-06-01T09:23:06","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=26149"},"modified":"2023-06-01T10:59:23","modified_gmt":"2023-06-01T09:59:23","slug":"uma-equacao-literal","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=26149","title":{"rendered":"Uma equa\u00e7\u00e3o literal"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_26149' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_26149' class='GTTabs_curr'><a  id=\"26149_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_26149' ><a  id=\"26149_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_26149'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Considera a equa\u00e7\u00e3o literal: \\[2a + 3\\left( {x &#8211; \\frac{1}{2}b} \\right) = \\frac{{a &#8211; x}}{3}\\]<\/p>\n<p>Se \\(a = 1\\) e \\(b = 2\\), temos:<\/p>\n<p><strong>[A]<\/strong> \\(x = \\frac{5}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[B]<\/strong> \\(x = \\frac{2}{5}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[C]<\/strong> \\(x = \\frac{3}{7}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[D]<\/strong> \\(x = \\frac{7}{3}\\)<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_26149' onClick='GTTabs_show(1,26149)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_26149'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>Considera a equa\u00e7\u00e3o literal: \\[2a + 3\\left( {x &#8211; \\frac{1}{2}b} \\right) = \\frac{{a &#8211; x}}{3}\\]<\/p>\n<p>Se \\(a = 1\\) e \\(b = 2\\), temos:<\/p>\n<p><strong>[A]<\/strong> \\(x = \\frac{5}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[B]<\/strong> \\(x = \\frac{2}{5}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[C]<\/strong> \\(x = \\frac{3}{7}\\)\u00a0 \u00a0 \u00a0 \u00a0 <strong>[D]<\/strong> \\(x = \\frac{7}{3}\\)<\/p>\n<\/blockquote>\n<p>\u00a0<\/p>\n<p>Substituindo, na equa\u00e7\u00e3o literal, os valores de <em>a<\/em> e de <em>b<\/em> indicados, vem: \\[\\begin{array}{*{20}{l}}{2 \\times 1 + 3\\left( {x &#8211; \\frac{1}{2} \\times 2} \\right) = \\frac{{1 &#8211; x}}{3}}&amp; \\Leftrightarrow &amp;{2 + 3\\left( {x &#8211; 1} \\right) = \\frac{{1 &#8211; x}}{3}}\\\\{}&amp; \\Leftrightarrow &amp;{6 + 9\\left( {x &#8211; 1} \\right) = 1 &#8211; x}\\\\{}&amp; \\Leftrightarrow &amp;{9x + x = &#8211; 6 + 9 + 1}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{2}{5}}\\end{array}\\]<\/p>\n<p>Portanto, a op\u00e7\u00e3o correta \u00e9 <strong>[B]<\/strong> \\(x = \\frac{2}{5}\\).<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_26149' onClick='GTTabs_show(0,26149)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Considera a equa\u00e7\u00e3o literal: \\[2a + 3\\left( {x &#8211; \\frac{1}{2}b} \\right) = \\frac{{a &#8211; x}}{3}\\] Se \\(a = 1\\) e \\(b = 2\\), temos: [A] \\(x = \\frac{5}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14061,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,714],"tags":[424,192],"series":[],"class_list":["post-26149","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-literais-e-sistemas","tag-8-o-ano","tag-equacoes-literais"],"views":66,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat06.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/26149","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=26149"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/26149\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14061"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=26149"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=26149"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=26149"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=26149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}